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Solve $$\frac{a_1}{1!}-\frac{a_2}{2!}+\frac{a_3}{3!}=\frac{1}{3}$$ where $a_1,a_2,a_3$ are positive integers.

By trial and error, I found $a_1=1$, $a_2=5$, $a_3=11$.

I ask if there are others solutions. Thanks.

Rodrigo de Azevedo
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SADIK MOHAMMED
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3 Answers3

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The equation is equivalent to $$6a_{1}-3a_{2}+a_{3}=2 $$ and there are infinite solution. It is sufficient to take $a_{3} $ such that $3\mid a_{3}-2 $ and $a_{2}=\frac{6a_{1}+a_{3}-2}{3} $. For example $\left(1,1,11\right)$, $\left(2,8,14\right)$ and so on.

Marco Cantarini
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All

$$a_3=2-6a_1+3a_2>0,$$ where $a_1,a_2>0$ are solutions.

There is a double infinity of them (in particular $(2n,m,2)$ with $m\ge n$).

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The equation is equivalent to $6 a_1 - 3 a_2 + a_3 = 2$ one solution (as just pointed out by A--B) $k,2k,2$. Other choices include taking $a_2 = n$, then you choose $a_1, a_3$ st $6a_1 + a_3 = 2 + 3n$. In fact, there are lots of solutions.

jim
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  • what about $1,3,5$ or $1,5,11$ – SADIK MOHAMMED Aug 17 '16 at 09:21
  • I dont agree, (1,3,5) is a solution and is not of the form (k,2k,2). – Jean Marie Aug 17 '16 at 09:27
  • Another independant basic family of solutions appears to be $(1,\ell,3\ell-4)$. By combining (in the sense of linear combination) it with the other family $(k,2k,2)$, one should get all solutions because of the affine plane interpretation given upwards. – Jean Marie Aug 17 '16 at 09:33
  • I am sorry but I don't think that the sentence "In fact, there are lots of solutions." is very informative. – Jean Marie Aug 17 '16 at 09:42
  • I meant by that it is possible to set $a_1$ to a fixed number, then $a_3$ and meant it for an exercise for SADIK MOHAMMED to explore the other possibilities. – jim Aug 17 '16 at 12:28